p varies directly as q and inversely as the square of r. Given p=9, q=6, r=3, find k. Then given p=12, q=72, r=a, find a.

Given $p=9$, $q=6$, $r=3$.
Substitute these values into the equation:

$$9 = k \frac{6}{3^2}$$ $$9 = k \frac{6}{9}$$ $$9 = k \frac{2}{3}$$

Multiply both sides by $\frac{3}{2}$:

$$k = 9 \times \frac{3}{2}$$ $$k = \frac{27}{2}$$ $$k = 13.5$$

The value of K is $\boxed{\frac{27}{2}}$ or $\boxed{13.5}$.

$p=12$, $q=72$, $r=a$.
Substitute these values and the calculated $k = \frac{27}{2}$ into the equation $p = k \frac{q}{r^2}$:

$$12 = \frac{27}{2} \frac{72}{a^2}$$

$$12 = \frac{27 \times 72}{2a^2}$$ $$12 = \frac{1944}{2a^2}$$ $$12 = \frac{972}{a^2}$$

Multiply both sides by $a^2$:

$$12a^2 = 972$$

Divide by 12:

$$a^2 = \frac{972}{12}$$ $$a^2 = 81$$

Take the square root of both sides:

$$a = \pm\sqrt{81}$$ $$a = \pm 9$$

The value of a is $\boxed{\pm 9}$.

$p=b$, $q=42$, $r=3$.
Substitute these values and the calculated $k = \frac{27}{2}$ into the equation $p = k \frac{q}{r^2}$:

$$b = \frac{27}{2} \frac{42}{3^2}$$

$$b = \frac{27}{2} \frac{42}{9}$$ $$b = \frac{27}{2} \times \frac{14}{3}$$ $$b = \frac{27 \times 14}{2 \times 3}$$ $$b = \frac{378}{6}$$ $$b = 63$$

The positive value of b is $\boxed{63}$.